CH3COOH(aq) +
H2O(l) <-----> CH3COO^-(aq) + H3O^+(aq)
initial 0.1
MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0 MÂ Â Â Â Â Â Â Â Â 0
M
change   Â
x                            Â
x          Â
x
equilibrium 0.1-x
MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
x          Â
x
Ka = [CH3COO-][H3O+]/[CH3COOH]
ka of acetic acid = 1.8*10^-5
1.8*10^-5 = X^2/(0.1-X)
X = 0.00133
So that,
at equilibrium
[H3O+] = X = 0.00133 M
pH = -log(H3O+)
  = -log0.00133
   = 2.876
Ka = CX^2
X = degree of dissociation = ?
C = inital concentration = 0.1 M
(1.8*10^-5) = 0.1*x^2
x = 0.0134
percentage dissociation = x*100
               Â
= 0.0134*100
              Â
= 1.34%
